Verified Solution • 2025–26 Session moderate

QJEEM21Maf86b: JEE Main 2021 Mathematics - Class 12 JEEM21Maf86b Solution: The value of , where [ x ] denotes the greatest integer x is :

The value of
lim⁡x→0+cos⁡−1(x−[x]2).sin⁡−1(x−[x]2)x−x3\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(x - {{[x]}^2}).{{\sin }^{ - 1}}(x - {{[x]}^2})} \over {x - {x^3}}} , where [ x ] denotes the greatest integer ≤\le x is :

Multiple Choice Options

  • A π\pi
  • B π4{\pi \over 4}
  • C π2{\pi \over 2}
  • D 0

Step-by-Step Verified Solution

lim⁡x→0+cos⁡−1(x−[x]2).sin⁡−1(x−[x]2)x−x3\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}\left( {x - {{[x]}^2}} \right).{{\sin }^{ - 1}}\left( {x - {{[x]}^2}} \right)} \over {x - {x^3}}}

=lim⁡x→0+cos⁡−1x1−x2.sin⁡−1xx= \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}x} \over {1 - {x^2}}}.{{{{\sin }^{ - 1}}x} \over x}

=cos⁡−10=π2= {\cos ^{ - 1}}0 = {\pi \over 2}

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